On Revenue Maximization with Sharp Multi-Unit Demands

Transcription

1 On Revenue Maximization with Sharp Multi-Unit Demands Ning Chen Xiaotie Deng Paul W. Goldberg Jinshan Zhang October 20, 2014 Abstract We consider markets consisting of a set of indivisible items, and buyers that have sharp multi-unit demand. This means that each buyer i wants a specific number d i of items; a bundle of size less than d i has no value. We consider the objective of setting prices and allocations in order to maximize the total revenue of the market maker. The pricing problem with sharp multi-unit demand buyers has a number of properties that the unit-demand model does not possess, and is an important question in algorithmic pricing. We consider the problem of computing a revenue maximizing solution for two solution concepts: competitive equilibrium and envy-free pricing. For unrestricted valuations, these problems are NP-complete; we focus on a realistic special case of correlated values where each buyer i has a valuation v i q j for item j, where v i and q j are positive quantities associated with buyer i and item j respectively. We present a polynomial time algorithm to solve the revenue-maximizing competitive equilibrium problem. For envy-free pricing, if the demand of each buyer is bounded by a constant, a revenue maximizing solution can be found efficiently; the general demand case is shown to be NP-hard. Keywords: Position Auction, Revenue Maximization, Envy-free, Competitive Equilibrium, Sharp Demand Division of Mathematical Sciences, Nanyang Technological University, Singapore. ningc@ntu.edu.sg. Department of Computer Science, Shanghai Jiao Tong University, China. dengxiaotie@gmail.com Department of Computer Science, University of Oxford, UK. Paul.Goldberg@cs.ox.ac.uk. Supported by EPSRC grant EP/G069239/1 Efficient Decentralised Approaches in Algorithmic Game Theory. Department of Computer Science, University of Liverpool, UK. Jinshan.Zhang@liv.ac.uk.

2 1 Introduction The problems considered in this paper are motivated by applications illustrated by the following examples. A publisher (e.g., a TV network) has some items (such as advertising slots) that are provided to potential customers (the advertisers). Each customer i has a demand that specifies the number of items that i needs. Given the demand requests from different customers, as well as the values that they are willing to pay, the problem that the publisher faces is how to allocate the items to customers at which prices. Demand is a practical consideration and has occurred in a number of applications. For instance, in TV (or radio) advertising [29], advertisers may request different lengths of advertising slots for their ads programs. In banner advertising, advertisers may request different sizes or areas for their displayed ads, which may be decomposed into a number of base units. A notable application of our model is where advertisers choose to display their advertisement using rich media (video, audio, animation) [4, 30] that would usually need a fixed number of positions while text ads would need only one position each. It has been formulated as a consecutive sharp-demand model in sponsored search in recent works [20, 14, 13]. Hatfield [26] studies mechanisms (that do not use money) in this context: each agent has a quota, a fixed number of items required, and has additive valuations for bundles of the items. Examples of this kind of situation include allocation of (multiple) projects to employees, items among heirs, allocating equipment time to scientists, tutorial sessions to students and players to sports teams. Our model contrasts with [26] in that we consider mechanisms with money (market prices arise), and our valuation functions are less general: for us, the matrix of buyer/item values has rank 1. We study the economic problem as a two-sided market where the supply side is composed of m indivisible items and each item j has a parameter q j, measuring the quality of the item. For example, in TV advertising, inventories of a commercial break are usually divided into slots of five seconds each, and every slot has an expected number of viewers. The other side of the market has n potential buyers where each buyer i has a demand d i (the number of items that i requests) and a value v i (the benefit to i for an item of unit quality). Thus, the valuation that i obtains from item j is given by v ij = v i q j. We suppose the valuation function is also additive by standard assumption in position auction. The v i q j valuation model has been considered by Edelman et al. [15] and Varian [32] in their seminal work for keywords advertising. We will focus on the sharp demand case, where every buyer requests exactly d i items. This scenario captures some similarity but is still quite different from single-minded buyers (i.e., each one desires a fixed combination of items) and is distinct from the relaxed demand case, where every buyer requests at most d i items. In the practical setting of rich media advertisement, one slot can be sold as a single text ad, or some given number of slots for one rich media ad. In practice, these slots would normally have to be adjacent. Here we do not explicitly impose that as a requirement, but it turns out to be satisfied by our solution, provided that slots are ordered by quality value. Given the valuations and demands from the buyers, the market maker decides on a price vector p = (p j ) for all items and an allocation of items to buyers, as an output of the market. The question is one of which output the market maker should choose to achieve certain objectives. In this paper, we assume that the market maker would like to maximize his own revenue, which is defined to be the total payment collected from the buyers. While revenue maximization is a natural goal from the market maker s perspective, buyers may have their own objectives as well. We aim to model a free market where consumers are price takers; thus, in a robust solution concept, one has to consider the performance of the whole market and the interests of the buyers. Competitive equilibrium provides such a solution concept that captures both market efficiency and fairness for the buyers. In a competitive equilibrium, every buyer obtains a best possible allocation 1

3 that maximizes his own utility and every unallocated item is priced at zero (i.e., market clearance). Competitive equilibrium is one of the central solution concepts in economics and has been studied and applied in a variety of domains [28]. Combining the considerations from the two sides of the market, an ideal solution concept therefore would be revenue maximizing competitive equilibrium. For sharp multi-unit demand buyers, when the valuations v ij are arbitrary, even determining the existence of a competitive equilibrium is NP-complete (see Appendix A). For our correlated valuation v i q j model, we have the following results. Theorem 1. For sharp multi-unit demand, a competitive equilibrium may not exist; even if an equilibrium is guaranteed to exist, a maximum equilibrium (in which each price is as high as it can be in any solution, see Definition 2.3 and Example 2.2) may not exist. Further, there is a polynomial time algorithm that determines the existence of an equilibrium, and computes a revenue maximizing one if it does. While (revenue maximizing) competitive equilibrium has a number of nice economic properties and has been recognized as an elegant tool for the analysis of competitive markets, its possible non-existence largely ruins its applicability. Such non-existence is a result of the market clearance condition required in the equilibrium (i.e., unallocated items have to be priced at zero). In most applications, however, especially in advertising markets, market makers are able to manage the amount of supplies. For instance, in TV advertising, publishers can freely adjust the length of a commercial break. Therefore, the market clearance condition becomes arguably unnecessary in those applications. This motivates the study of envy-free pricing (that is, envy-free item pricing [18]) which only requires the fairness condition in the competitive equilibrium, where no buyer can get a larger utility from any other allocation for the given prices. In contrast with competitive equilibrium, an envy-free solution always exists (a trivial one is obtained by setting all prices to ). Once again, taking the interests of both sides of the market into account, revenue maximizing envy-free pricing is a natural solution concept that can be applied in those marketplaces. The study of algorithmic computation of revenue maximizing envy-free pricing was initiated by Guruswami et al. [23], where the authors considered two special settings with unit demand buyers and single-minded buyers and showed that a revenue maximizing envy-free pricing is NP-hard to compute. Because envy-free pricing has applications in various settings and efficient computation is a critical condition for its applicability, there is a surge of studies on its computational issues since the pioneering work of [23], mainly focusing on approximation solutions and special cases that admit polynomial time algorithms, e.g., [25, 2, 7, 3, 6, 10, 19, 16, 9, 21]. The NP-hardness result of [23] for unit demand buyers implies that we cannot hope for a polynomial time algorithm for general v ij valuations in the multi-unit demand setting, even for the very special case when one has positive values for at most three items [9]. However, it does not rule out positive computational results for special, but important, cases of multi-unit demand. For v i q j valuations with multi-unit demand, where the hardness reductions of [23, 9] does not apply, we have the following results. Theorem 2. There is a polynomial time algorithm that computes a revenue maximizing envy-free solution in the sharp multi-unit demand model with v i q j valuations if the demand of every buyer is bounded by a constant. On the other hand, the problem is NP-hard if the sharp demand is arbitrary, even if there are only three buyers. For relaxed multi-unit demand, a standard technique can reduce the problem to the unit-demand version: each buyer i with demand d i can be replaced by d i copies of buyer i, each of whom requests one item. Note that under the sharp demand constraint, this trick is no longer applicable. 2

4 We summarize our results in the following table. Here, we have a complete overview of the existence and computation of competitive equilibrium and envy-free pricing with multi-unit demand buyers. Most of our results are positive, suggesting that competitive equilibrium and envy-free pricing are candidate solution concepts to be applicable in the domains where the valuations are correlated with respect to the quality of the items. Competitive equilibrium Envy-free pricing Unit demand (general values v ij ) existence yes [31] yes (trivial) max revenue computation P [31, 12] NP-hard [23] not always (P decidable) existence yes (trivial) Sharp multi-unit demand (NP-hard for general v ij ) (v ij = v i q j ) max revenue P (constant demand) computation P (if one exists) NP-hard (arbitrary demand) Table 1: Summary of previous work and our results. Despite the recent surge in the studies of algorithmic pricing, multi-unit demand models have not received much attention. Most previous work has focused on two simple special settings: unit demand and single-minded buyers, but arguably multi-unit demand has much more applicability. While the relaxed demand model shares similar properties to unit demand (e.g., existence, solution structure, and computation), the sharp demand model has a number of features that unit demand does not possess. Existence of equilibrium. In unit or relaxed demand case, the competitive equilibrium always exists, moreover, the maximum and minimum equilibrium (an equilibrium price vector no more than any equilibrium price vector in each coordinates) always exists. As discussed above, a competitive equilibrium may not exist in the sharp demand model. Further, even if a solution exist, the solution space may not form a distributive lattice (Any price vector between the minimum equilibrium price vector and maximum equilibrium price vector in each coordinate is an equilibrium price vector). Over-priced items. In unit or relaxed demand, the price p j of any item j is always at most the value v ij of the corresponding winner i. This no longer holds for sharp multi-unit demand. Specifically, even if p j > v ij, buyer i may still want to get j since his net profit from other items may compensate his loss from item j (see Example 2.3) 1. This property enlarges the solution space and adds an extra challenge to finding a revenue maximizing solution. Our Techniques. To compute a competitive equilibrium, we first find a candidate winner set (at least one optimal winner set is a candidate winner set, see Definition 4.1 for details), which can be proved to be an equilibrium winner set if a competitive equilibrium exists; then, with this set, we transform the computation of competitive equilibria to a linear program of exponential size, which can be solved by the ellipsoid algorithm in polynomial time. The situation becomes complicated when 1 This phenomenon does occur in our real life. For example, in most travel packages offered by travel agencies, they could lose money for some specific programs; but their overall net profit could always be positive. 3

5 finding an optimal envy-free solution. Actually, we prove that it is NP-hard to compute an optimal envy-free solution even if there are only three buyers. Hence, our efforts focus on the special, yet very important bounded-demand case. To compute an optimal envy-free solution for bounded demand, certain candidate winner sets (the number of such sets is polynomial) are defined and found; and crucially, there is at least one optimal winner set in our selected candidate winner sets. For each candidate winner set, if the demand is bounded by a constant, we present a linear programming to characterize its optimal solution when the allocation is known. Finally, a dynamic programming algorithm is provided to find the allocation sets when a candidate winner set is fixed. Both the linear programming and the dynamic programming run in polynomial time. 1.1 Related Work There are extensive studies on multi-unit demand in economics, see, e.g., [1, 17, 8]. Our study focuses on sharp demand buyers. An alternative model is when buyers have relaxed multi-unit demand (i.e., one can buy a subset of at most d i items), where it is well known that the set of competitive equilibrium prices is non-empty and forms a distributive lattice [31, 22]. This immediately implies the existence of an equilibrium with maximum possible prices; hence, revenue is maximized. Demange, Gale, and Sotomayor [12] proposed a combinatorial dynamics which always converges to a revenue maximizing (or minimizing) equilibrium for unit demand; their algorithm can be easily generalized to relaxed multi-unit demand. From an algorithmic point of view, the problem of revenue maximization in envy-free pricing was initiated by Guruswami et al. [23], who showed that computing an optimal envy-free pricing is APX-hard for unit-demand bidders and gave an O(log n) approximation algorithm. Briest [6] showed that given appropriate complexity assumptions, the unit-demand envy-free pricing problem in general cannot be approximated within O(log ɛ n) for some ɛ > 0. Hartline and Yan [24] characterized optimal envy-free pricing for unit-demand and showed its connection to mechanism design. For the multi-unit demand setting, Chen et al. [10] gave an O(log D) approximation algorithm when there is a metric space behind all items, where D is the maximum demand, and Briest [6] showed that the problem is hard to approximate within a ratio of O(n ɛ ) for some ɛ, unless NP ɛ>0 BP T IME(2nɛ ). For the problem of maximizing social welfare via truthful mechanisms, [27] investigates a similar sharp demand model for combinatorial auctions in the case that bidders all have the same demand d. [27] obtain positive algorithmic results for approximation. Recent work by Feldman et al. [18] studies envy-free revenue maximization problem with budget but without demand constraints and presents a 2-approximate mechanism for the envy-free pricing problem. Another line of research is on single-minded bidders, including, for example, [23, 3, 2, 7, 11, 16]. To the best of our knowledge, this paper is the first to study algorithmic computation of sharp multi-unit demand. Most related work to this is a follow-up paper [5], where the authors prove that, based on our work, the maximum revenue of envy-free solutions for v i q j valuation with sharp multi-unit demand cannot be approximated within a factor O(m 1 ɛ ) for arbitrary demands, for any ɛ > 0, unless P = NP, and provide a simple O(m)-approximation algorithm. [5] also studies an interesting subclass of proper instances and gives a tight 2-approximation algorithm for this class. Deng et al. [13] investigates a consecutive demand variant in which items are arranged in a sequence and buyers want items that are consecutive in the sequence. 4

6 2 Preliminaries 2.1 Settings and Definitions We have a market with m indivisible items, M = {1, 2,..., m}, where each item j has unit supply and a parameter q j > 0, representing the quality or desirability of j. In the market, there are also n potential buyers, N = {1, 2,..., n}, where each buyer i has a value v i > 0, which gives the benefit that i obtains for each unit of quality. Hence, the valuation that buyer i has for item j is v ij = v i q j. We suppose the valuation function is also additive. In addition, each buyer i has a demand request d i Z +, which specifies the number of items that i would like to get. We assume that d i is a sharp constraint, i.e., i gets either exactly d i items 2 or nothing at all. Our model therefore defines a market with multi-unit demand buyers and unit supply items. For any subset of buyers S N, we use d(s) = i S d i to denote the total demand of items by buyers in S. An outcome of the market is a tuple (p, X), where p = (p 1,..., p m ) 0 is a price vector, where p j is the price charged for item j; X = (X 1,..., X n ) is an allocation vector, where X i is the set of items that i wins. If X i, we say i is a winner and have X i = d i due to the demand constraint; if X i =, i does not win any items and we say i is a loser. Further, since every item has unit supply, we require X i X i = for any i i. If j X i, we use i = b(j) to represent the buyer of j M. Given an output (p, X), let u i (p, X) denote the utility of i. That is, if X i, then u i (p, X) = j X i (v ij p j ); if X i =, then u i (p, X) = 0. Definition 2.1 (Envy-freeness). We say a tuple (p, X) is an envy-free solution if every buyer is envy-free, where a buyer i is envy-free if the following conditions are satisfied: if X i, then (i) u i (p, X) 0, and (ii) for any other subset of items T with T = d i, u i (p, X) (v ij p j ); j T if X i = (i.e., i wins nothing), then, for any subset of items T with T = d i, (v ij p j ) 0. Envy-freeness captures fairness in the market the utility of everyone is maximized at the corresponding allocation for the given prices. That is, if i wins a subset X i, then i cannot obtain a higher utility from any other subset of the same size; if i does not win anything, then i cannot obtain a positive utility from any subset with size d i. It is easy to see that an envy-free solution always exists (e.g., set all prices to be and allocate nothing to every buyer). Another solution concept we will consider is competitive equilibrium, which requires that, besides envy-freeness, every unsold item must be priced at zero (or at any given reserve price). Such market clearance condition captures efficiency of the whole market. The formal definition is given below. Definition 2.2 (Competitive equilibrium). We say a tuple (p, X) is a competitive equilibrium if it is envy-free, and for each item j, p j = 0 if no-one wins j in the allocation X. 2 By the nature of the solution concepts considered in the paper, it can be assumed without loss of generality that i will not get more than d i items. j T 5

7 For a given output (p, X), the revenue collected by the market maker is defined as n i=1 j X i p j. Note that by the definition of competitive equilibrium, the revenue collected from a competitive equilibrium is m i=1 p j. We are interested in revenue maximizing solutions, specifically, revenue maximizing competitive equilibrium (if one exists) and revenue maximizing envy-free pricing. The main objective of the paper is algorithmic computations of these two optimization problems. To simplify the following discussions, we sort all buyers and items in non-increasing order of their unit values and qualities, respectively, i.e., v 1 v 2 v n and q 1 q 2 q m. Let K be the number of distinct values in the set {v 1,..., v n }. Let A 1,..., A K be a partition of all buyers where each A k, k = 1, 2,..., K, contains the set of buyers that have the kth largest value. 2.2 Examples It is well known that a competitive equilibrium always exists for unit demand buyers (even for general v ij valuations) [31]; for our sharp multi-unit demand model, however, a competitive equilibrium may not exist, as the following example shows. Example 2.1 (Competitive equilibrium need not exist). There are two buyers i 1, i 2 with values v i1 = 10 and v i2 = 9, and demands d i1 = 1 and d i2 = 2, respectively, and two items j 1, j 2 with unit quality q j1 = q j2 = 1. If i 1 wins an item, without loss of generality, say j 1, then j 2 is unsold and p j2 = 0; by envy-freeness of i 1, we have p j1 = 0. Thus, i 2 envies the bundle {j 1, j 2 }. If i 2 wins both items, then p j1 + p j2 v i2 j 1 + v i2 j 2 = 18, implying that p j1 9 or p j2 9; thus, i 1 is not envy-free. Hence, there is no competitive equilibrium in the given instance. In the unit demand case, it is well-known that the set of equilibrium prices forms a distributive lattice; hence, there exist extremes which correspond to the maximum and the minimum equilibrium price vectors. In our multi-unit demand model, however, even if a competitive equilibrium exists, maximum equilibrium prices may not exist. Definition 2.3 (Maximum Equilibrium). A price vector p is called a maximum equilibrium price vector if for any other equilibrium price vector q, p j q j for every item j. An equilibrium (p, X) is called a relaxed maximum equilibrium if p is a maximum price vector. Example 2.2 (Maximum equilibrium need not exist). There are two buyers i 1, i 2 with values v i1 = 10, v i2 = 1 and demands d i1 = 2, d i2 = 1, and two items j 1, j 2 with unit quality q j1 = q j2 = 1. It can be seen that allocating the two items to i 1 at prices (19, 1) or (1, 19) are both revenue maximizing equilibria; but there is no equilibrium price vector which is at least both (19, 1) and (1, 19). 2.3 Over-Priced Items Because of the sharp multi-unit demand, an interesting and important property is that it is possible that some items are over-priced ; this is a significant difference between sharp multi-unit and unit demand models. Formally, in a solution (p, X), we say an item j is over-priced if there is a buyer i such that j X i and p j > v i q j. That is, the price charged for item j is larger than its contribution to the utility of its winner. Example 2.3 (Over-priced items in a revenue maximizing solution). There are two buyers i 1, i 2 with values v i1 = 20, v i2 = 10 and demands d i1 = 1 and d i2 = 2, and three items j 1, j 2, j 3 with qualities q j1 = 3, q j2 = 2, q j3 = 1. We can see that the allocations X i1 = {j 1 }, X i2 = {j 2, j 3 } and prices (45, 25, 5) constitute a revenue maximizing envy-free solution with total revenue 75, where item j 2 is over-priced. If no items are over-priced, the maximum possible prices are (40, 20, 10) with total revenue 70. 6

8 We have the following characterization for over-priced items in an equilibrium solution. Lemma 2.1. For any given competitive equilibrium (p, X), the following claims hold: If there is any unallocated item, then there are no over-priced items. At most one winner can have over-priced items; further, that winner, say i, must be the one with the smallest value among all winners in the equilibrium (p, X). That is, for any other winner i i, we have v i > v i. Proof. The first claim is obvious since any unallocated item j is priced at 0; thus if there is a winner i and item j X i such that p j > v i q j, then i would envy the subset X i {j } \ {j}. To prove the second claim, suppose there are two winners i, i where v i v i, and suppose that i has over-priced item j. Since i is envy-free, his own utility must be non-negative; we know there is an item j X i such that v i q j p j. This implies that v i q j p j ; thus, i would envy the subset X i {j } \ {j}, a contradiction. 2.4 Properties We present some observations regarding envy-freeness and competitive equilibrium. Our first observation implies that a winner is envy-free if and only if he prefers each of his allocated items to any other item. Lemma 2.2. Given any solution (p, X) and any winner i, if i is envy-free then v ij p j v ij p j for any items j X i and j / X i. On the other hand, if i is not envy-free, then there is j X i and j / X i such that v ij p j < v ij p j. Proof. If i is envy-free but (for j X i and j / X i ) v ij p j < v ij p j, it is easy to see that i would envy subset X i {j } \ {j}, a contradiction. If i is not envy-free, then there is a subset T of items with T = d i such that j X i (v ij p j ) < j T (v ij p j ). Since X i = T, the inequality holds for at least one item, i.e., there is j X i and j / X i such that v ij p j < v ij p j. Lemma 2.3. For any envy-free solution (p, X), suppose there are two buyers i, i with values v i > v i and two items j and j that are allocated to i and i respectively, i.e., j X i and j X i. Then q j q j. Proof. By the above Lemma 2.2, we have v i q j p j v i q j p j v i q j p j v i q j p j Adding the two inequalities together, we get (v i v i )(q j q j ) 0, yielding the desired result. Lemma 2.3 implies that in any envy-free solution, the allocation of items is monotone in terms of their amount of qualities and the values of the winners, i.e., winners with larger values win items with larger qualities. However, it does not imply that the value of every winner is larger than or equal to the value of any loser. For instance, consider three buyers i 1, i 2, i 3 and two items j 1, j 2 with q j1 = 2 and q j2 = 1. The values and demands are v i1 = 1.3, v i2 = 1, v i3 = 0.9 and d i1 = 1, d i2 = 2, d i3 = 1. Then prices p j1 = 2.2, p j2 = 0.9 and allocations X i1 = {j 1 }, X i2 =, X i3 = {j 2 } constitute a revenue maximizing envy-free solution. In this solution, v i2 > v i3, but i 2 does not win any item (because of the sharp demand constraint) whereas i 3 wins item j 2. 7

9 Lemma 2.4. If there is a competitive equilibrium (p, X), then for any winner i, item j X i and unallocated item j, we have q j q j. Proof. Since item j is not allocated to any buyer, its price p j = 0. By envy-freeness and Lemma 2.2, we have v i q j v i q j p j v i q j p j = v i q j, which implies that q j q j. By the above characterization, in any competitive equilibrium, all allocated items have larger qualities. Hence, by Lemmas 2.3 and 2.4, we know that if the set of winners is fixed in a competitive equilibrium, the allocation is determined implicitly as well. On the other hand, we observe that Lemma 2.4 does not hold if (p, X) is an (revenue maximizing) envy-free solution. For instance, consider two buyers i 1, i 2 with values v i1 = 10, v i2 = 1 and demand d i1 = 1, d i2 = 10, and twelve items j 1, j 2,..., j 12 with qualities q j1 = 10, q j2 = 5, q j3 = = q j12 = 1. It can be seen that in the optimal envy-free solution, we set prices p j1 = 91, p j2 =, p j3 = = p j12 = 1, and allocate X i1 = {j 1 }, X i2 = {j 3,..., j 12 }, which generates total revenue = 101. In this solution, q j2 > q j3 = = q j12, but item j 2 is not allocated to any buyer. Lemma 2.5. Given an envy-free solution (p, X), a loser l and any subset T of d l items, the following property cannot hold: A non-empty subset of items in T are either allocated to winners with values smaller than v l or priced at 0; any other elements of T are allocated to winners having the same value v l as l. Note that this is a result about envy-free prices, not just competitive equilibrium. Proof. Let (p, X) be an envy-free pair of price and allocation vectors. Given the loser l and T satisfying the conditions of the statement of the Lemma, we show how to construct a set T of items that l envies. Let T = T 0 T 1 T s be a partition of T where T 0 consists of items priced at 0 in (p, X) and for i > 0, T i = T X i and s is the number of non-empty elements in {T X i, i [n]}. Note that any non-empty T i satisfies v i v l, and if T 0 = then T i for some i > 0 with v i < v l. Note that T 0 satisfies j T 0 v i q j p j 0, where the inequality is strict if T 0 is non-empty. Let T 0 = T 0. Consider any non-empty T i (with i > 0). Let T i be the T i items j X i that maximize v i q j p j. We have j T i v iq j p j 0. Hence j T i v lq j p j 0, with strict inequality if v i < v l. Summing these inequalities, we have s i=0 j T i v lq j p j 0, and in fact the inequality is strict since at least one of the s + 1 inequalities is strict. Let T = T 0 T 1 T s; T = T = d l and we have shown that l envies T. 3 Computation of Competitive Equilibrium Our main result of this section is the following. Theorem 3.1. There is a polynomial algorithm to determine the existence of a competitive equilibrium; and if one exists, it computes a revenue maximizing equilibrium. Thus, both the existence problem and the maximization problem become tractable, as a result of the correlated valuations v ij = v i q j. The algorithm, called Max-CE, is divided into two steps. The first step is to compute a set of candidate winners if an equilibrium exists. The second step is to calculate a candidate equilibrium and verify if it is indeed a (revenue maximizing) equilibrium. Recall that A k, 1 k K denotes all the buyers with the kth largest value. 8

10 Max-CE stage Let S be the set of candidate winners 2. Let k 1 and let D m be the number of available items 3. While k K If d i > D for every i A k, let k k + 1 Else Let S = {i i A k, d i D} If d(s) > D (a) If there is S S s.t. d(s ) = D let S S S, and goto Max-CE stage 2 (b) Else, a competitive equilibrium does not exist, and return Else d(s) D (c) Let S S S, D D i S d i, k k Goto Max-CE stage 2 Note that in the above step 3(a) we check whether there is S S such that d(s ) = D; this is equivalent to solving a subset sum problem. However, in our instance, each demand satisfies d i m. Hence, a dynamic programming approach can solve the problem in time O(n 2 m). Hence, stage 1 runs in strongly polynomial time. An input to Max-CE is all the n buyers with valuation v i and demand d i and all the m items with qualities q j. Lemma 3.1. If an input to Max-CE has a competitive equilibrium (p, X), then stage 1 will not return that an equilibrium does not exist at step 3(b). Proof. Let (p, X) be a competitive equilibrium of an input to Max-CE. In this proof, when we refer to winning/losing buyers, or allocated/unallocated items, we mean with respect to (p, X). In particular, let W be the set of winners of (p, X). Suppose that Max-CE stage 1 exits on the k-th iteration of the loop. We claim that during the first k 1 iterations, all buyers added to S must be winners i.e. S W. To see this, suppose alternatively that at iteration k < k, buyer l is the first loser to be added to S. In that case, l has d l items that satisfy the conditions of Lemma 2.5, contradicting envy-freeness (Suppose that the winners found by the algorithm during the first k 1 iterations are given their allocation in (p, X). At iteration k, the algorithm has more than d l available items, some of which are allocated to buyers with value less than l, or are unallocated.). Second, we claim that S = W ( k 1 i=1 A i). We will inductively shows that W ( k 1 i=1 A i) S. The base case k = 2, since step 3(c) is executed during the first k 1 iterations, each i A 1 with d i m will be added to S, which gives that W A 1 S. Suppose now k 2 is true, we argue that the case k 1 is true. Since step 3(c) is executed during the first k 1 iterations, each buyer i A k 1 with d i D will be added to S. This is the maximum number of buyers in A k 1 which can be added into W. Therefore, W ( k 1 i=1 A i) S. Since S W ( k 1 i=1 A i), the claim S = W ( k 1 i=1 A i) holds. At the final iteration k we must have S (otherwise the algorithm will begin a new iteration). Since d(s) > D, we have S\W (members of S have too much demand for them all to be able to win). Since there is no subset S S such that d(s ) = D, we have d(s W ) < D. Hence, there 9

11 are items that are not allocated to buyers in S (S W ). Note that these items are either priced 0 or allocated to winners with values smaller than v k. Since S = W ( k 1 i=1 A i), there are D items allocated to winners with values no more than v k or priced 0, among which there are items either priced 0 or allocated to winners with values smaller than v k. Hence, for any loser i S \ W where d i < D, we can find d i items that satisfy the condition of Lemma 2.5: a contradiction. Lemma 3.2. A revenue maximizing competitive equilibrium (p, X) can be converted to one with equal revenue whose winning set is S. Proof. Assume that the given instance has a competitive equilibrium (p, X) and that Max-CE enters Max-CE stage 2 at the kth iteration with the set of candidates S. Let W be the set of winners of (p, X), and let W = W (A 1 A k 1 ) and W = W \ W. Let S 1 = S (A 1 A k 1 ) and S 2 = S \S 1 (note that S 2 A k ). From the analysis of the above lemma and Lemma 2.5, we know that (i) W = S 1, (ii) W A k, and (iii) d(w ) = d(s 2 ). (i) is proved in Lemma 2.5. For (ii), if W \A k, then W A k will be selected by Max-CE and Max-CE in stage 1 will enter k th iteration with k > k, which contradicts that k is the final iteration in stage 1 of Max-CE. Hence, W A k. For (iii), in kth iteration of stage 1, if Max-CE enters step 3(a), then d(s 2 ) = m d(s 1 ) = m d(w ) d(w ). Suppose d(s 2 ) > d(w ), which means that some buyer in S 2 will be a loser in (p, X). Due to (ii), in (p, X), there are m d(w ) items priced 0 and d(w ) items allocated to W. Hence, the loser in S 2 of (p, X) will not be envy-free by Lemma 2.5, a contradiction. If in kth iteration of stage 1, Max-CE enters step 3(c), then S 2 will be winners by previous argument and we have d(w ) = d(s 2 ). Thus, the only difference between S and W lies in the selection of buyers in A k (this is due to possibly multiple choices in step 3(a) in Max-CE stage 1). Due to envy-freeness, we have i W \S 2 u i (p, X) = i W \S 2 j X i (v i q j p j ) 0 i S 2 \W u i (p, X) Since all buyers in W \ S 2 and S 2 \ W have the same value, we know that the above inequalities are tight. Thus, if we reassign the items in i W X i to the buyers in S 2 and keep the same prices, the resulting output will still be an equilibrium. Given the above characterization, the second step of the algorithm Max-CE is described in Max- CE stage 2. In the LP of Max-CE stage 2, there are m variables where each item j has a variable p j. The first two constraints ensure that the price vector is a set of feasible market clearing prices. The third condition guarantees that all winners are envy-free. The last condition says that for each loser i and any subset of items T with T = d i, i cannot obtain a positive utility from T. Notice that it is possible that there are exponentially many combinations of T ; thus the LP has an exponential number of constraints. However, observe that for any given solution p, it is easy to verify if p is a feasible solution of the LP or find a violated constraint. In particular, for every loser i, we can order all items j in decreasing order of v i q j p j and verify the subset T composed of the first d i items; if i cannot obtain a positive utility from such T, then i is envy-free. Therefore, there is a separation oracle to the LP, and thus, the ellipsoid method can solve the LP in polynomial time. Hence, the total running time of Max-CE is polynomial. 10

12 Max-CE stage Allocation X is constructed as follows: Let Xi, for each buyer i / S For each i S in non-increasing order of v i allocate d i of the remaining items to i in non-increasing order of q j 6. Price p is computed according to the following linear program: max i S j X p i j s.t. p j 0 j p j = 0 j / i S X i v i q j p j v iq j p j j T (v iq j p j ) 0 i S, j X i, j / X i i / S, T with T = d i 7. If the above linear program has a feasible solution, output the tuple (p, X ) 8. Else, return that a competitive equilibrium does not exist If the algorithm returns a tuple (p, X ), certainly it is a competitive equilibrium; further, it is a revenue maximizing equilibrium because of the objective function in the LP. It is therefore sufficient to show the following claim to complete the proof of Theorem 3.1. Lemma 3.3. If there exists a competitive equilibrium, then stage 2 will not claim that an equilibrium does not exist at step 8. Proof. If there is a competitive equilibrium (p, X), let W be the set of winners of the equilibrium. By Lemma 3.1, we know that Max-CE will enter Max-CE stage 2. By the above discussions, we know that W and S only differ in the last kth iteration of the main loop of Max-CE stage 1 and replacing all winners in W A k with S A k gives an equilibrium as well. Further, by Lemma 2.3 and 2.4, the allocation of items to the winners in W is fixed. Hence, the equilibrium price vector p gives a feasible solution to the LP in the stage 2, which implies that the algorithm will not claim that an equilibrium does not exist. 4 Computation of Envy-Free Pricing In this section, we will ignore the market clearance condition (i.e. that unsold items are priced at 0) and only consider envy-freeness. We noted earlier that an envy-free solution always exists. Our main results are the following. Theorem 4.1. For the sharp multi-unit demand with v i q j valuations, it is NP-hard to solve the revenue-maximizing envy-free pricing problem, even if there are only three buyers. However, if the demand of each buyer is bounded by a constant, then the revenue-maximizing envy-free pricing problem can be solved in polynomial time. We note that the correlated v i q j valuations are crucial to derive an efficient computation when the demands are bounded by a constant; in contrast, for unit-demand, the envy-free pricing is NP-hard for general valuations v ij even if every buyer is interested in at most three items [9]. 11

13 4.1 Algorithm for Constant Demands Throughout this subsection, let be a constant where d i for any buyer i, and again, buyers and items are sorted according to their values and qualities. For any tuple (p, X), we assume that all unsold items are priced at. This assumption is without loss of generality for envy-freeness. We will first explore some important properties of an (optimal) envy-free solution, then at the end of the section present our algorithm Candidate Winner Sets For a given set S of buyers, let max(s) and min(s) denote the buyer in S that has the largest and smallest index, respectively. Definition 4.1 (Candidate winner set). Given a subset of buyers S, let k = max{r A r S }. We say S is a candidate winner set if the total demand of buyers in S is at most m, i.e., d(s) m, and for any i A 1 A k 1 \ S, d i > d i. i S: i >i The definition of candidate winner set is closely related to envy-freeness. Indeed, due to Lemma 2.5, the above definition defines a slightly larger set (than all possible sets of winners in envy-free solutions) as the inequality does not consider all the buyers completely in the same value category v j. However, this definition makes it easier for us to describe and analyze the algorithm. Proposition 4.1. For any envy-free solution (p, X), let S = {i X i } be the set of winners. Then S is either a candidate winner set or S =. Proof. The claim follows directly from Lemma 2.5. By Proposition 4.1, in order to calculate the optimal envy-free solutions, we need only to compute the envy-free solutions whose winner sets are candidate winner sets. Therefore, Algorithm FindWinners(S) is a procedure for finding candidate winner sets based on Lemma 2.5 and Proposition 4.1. It is an inductive procedure where the buyer with larger value must be selected as a winner if his demand is no more than the total demands of all the winners with smaller values (otherwise by Lemma 2.5, this buyer will be a loser and not be envy-free if he is not selected). FindWinners(S): Input a set of buyers S Let i max = max(s) and assume i max A k Initially let W S = S For each buyer j A 1 A k 1 in reverse order If j / S and d j d i, let W S W S {j} i W S : i>j Return W S Proposition 4.2. For any subset of buyers S, let W S =FindWinners(S). If d(w S ) m, then W S is a candidate winner set and for any candidate winner set S S, W S S. If d(w S ) > m, then there is no candidate winner set containing S. 12

14 Proof. Obviously, if d(w S ) m, then from the definition of candidate winner set, we know W S is a candidate winner set. Still, by the definition of candidate winner set, for any j in W S \S, any candidate winner set S S, since d j d i, then d j d i (since S S), thus, j S, hence, i W S : i>j W S S. Therefore, the second statement follows. i S : i>j Similar to FindWinners(S), FindLoser(S) is also an inductive procedure based on the observation that if a loser is envy-free then a loser with the same valuation but with a larger demand will also be envy-free. For more details, see the proof of Proposition 4.3. FindLoser(S): Input a candidate winner set S Let i min = min(s) and assume i min A j Initially let L S =, and α = For each k = j, j + 1,..., K Let i 0 = arg min{d i i A k \ S} If d i0 < α, let L S L S {i 0 } and α d i0 Return L S Proposition 4.3. For any given tuple (p, X) with winner set S, suppose that S is a candidate winner set and let L S =FindLoser(S). If all losers in L S are envy-free with respect to (p, X), then all other losers are envy-free as well. Proof. Assume there exists a loser i who is not envy-free, that is, such that there exists a set T of d i items such that j T (v i q j p j ) > 0. This implies that there exists T T with T = d i such that j T (v iq j p j ) j T (v i q j p j ) > 0: a contradiction. Hence, by the rules of FindLoser, we know that if all the losers in L S are envy-free, all other losers in A j A K are envy-free as well. On the other hand, for any loser j A 1 A j 1, since S is a candidate winner set, we know that d j > d i = d(s). Since all unsold items are priced at i S: i>j, we know that j is envy-free. Hence, all losers are envy-free Bounding the Number of Candidate Winner Sets We have the following bound on the number of candidate winner sets. Lemma 4.1. For any k {2,..., K} and S A k, suppose d(s) m. Let C = { S S S A 1 A k 1 and S S is a candidate winner set } Then C m d(s). Proof. Let a = d(s) and l be the index of the buyer max(a k 1 ). We add buyers l, l 1, l 2,..., 1 into S in sequence and maintain all the possible candidate winner sets. Let C 0 = {S}. In general, we have constructed C t containing all the candidate winner sets of {l, l 1, l 2,..., l t + 1} S. We order C t = {S t,1, S t,2,..., S t,nt } such that d(s t,1 ) d(s t,2 ) d(s t,nt ) m. We will inductively prove that d(s t,i ) id(s), for t = 0, 1,, l. The base case t = 0 is trivial since C 0 = {S}. Suppose the claim holds for some other value of t. That is, we have constructed C t containing all the candidate winner sets of {l, l 1, l 2,..., l t+1} S 13

15 with d(s t,i ) id(s), for any i n t. Now for the case t + 1, which means we will add l t into C t to construct C t+1. Let t s = max{i : d(s t,i ) < d l t } if {i : d(s t,i ) < d l t }, otherwise t s = 0. Let S t+1,j = S t,j for j = 1, 2,, t s, S t+1,j+ts = S t,j {l t} for j = 1, 2,..., n t. Clearly d(s t+1,i ) id(s) for i t s by the inductive hypothesis. Also, d(s t+1,j+ts ) = d(s t,j ) + d l t jd(s) + d(s t,ts ) (j + t s )d(s). Let n t+1 = max{i : d(s t+1,i ) m}. Clearly the claim follows for the case t + 1. The lemma follows by the condition m d(s l,nl ) n l d(s) Optimal Winner Sets Definition 4.2 (Optimal winner set). A subset of buyers S is called an optimal winner set if there is a revenue maximizing envy-free solution (p, X) such that S is its set of winners. Proposition 4.4. Let S be an optimal winner set and let k = max{r A r S }. For any S A k, if d(s ) = d(s A k ), then S (S\A k ) is an optimal winner set as well. Before proving the proposition, we first establish the following claim. Claim 4.1. Let (p, X) be a revenue-maximizing envy-free solution and let S be the winning set in (p, X), and let k = max{r A r S }. Then every buyer in A k has utility zero. Proof. Of course, every loser in A k has utility zero. To show that every winner in A k has utility zero, we show that if there exists a winner who has positive utility, then prices can be raised to the point where his utility becomes zero, while maintaining envy-freeness (contradicting the assumption that (p, X) maximizes revenue). Let i max be the buyer in A k S with the highest utility. Let δ = u i max (p, X) d imax. We claim that (p + δ, X) is an envy-free solution as well, where the price of each item is increased by δ. Obviously we have δ 0, and the conclusion holds trivially if δ = 0. Suppose δ > 0. For the tuple (p + δ, X), since all items have their prices increased by the same amount, all losers are still envy-free and all winners would not envy the items they don t get. Hence, we need only to check that each winner still gets a non-negative utility. For i max, we have u imax (p + δ, X) = 0. For any other winner i i max, it holds that v i v imax. Since i does not envy any item in (p, X), for any item j X i and j X imax, it holds that v i q j p j v i q j p j, hence, p j v i (q j q j ) + p j. Then, we get j X p j imax (v i (q j q j ) + p j ) j X = v i q j imax (v i q j p j ). d imax d imax This implies that p j + δ v i q j. Hence, u i (p + δ, X) = j X i (v i q j p j δ) 0. Therefore, (p + δ, X) is an envy-free solution. We are now ready for the proof of Proposition

16 Proof of Proposition 4.4. Since S is an optimal winner set, there is an optimal envy-free solution (p, X) such that S = {i X i }. We construct a new allocation X with winner set S (S\A k ) as follows: For any i / A k, X i = X i. For any i A k \ S, X i =. For all the buyers in S, allocate items in i S A k X i to them arbitrarily. (The allocation is feasible as d(s ) = d(s A k ).) Obviously, (p, X ) generates the same revenue as (p, X). We claim that (p, X ) is an envy-free solution (this implies our desired result that S (S\A k ) is an optimal winner set). For any buyer i / A k, since prices are not changed, i is still envy-free. Next we prove that all buyers i A k are envy-free in (p, X ). Let J = i S Ak X i be the set of items allocated to buyers in A k ; we also have J = i S X i. Suppose first that S A k = S = 1; in this case (p, X) differs trivially from (p, X ), so (p, X ) is envy-free. Alternatively, there is some buyer ī A k with d ī < d(s A k ). We show that any item j J allocated to any buyer i A k in (p, X ), affords zero utility to i, i.e. j satisfies v i q j = p j. Let v be the value shared by all i A k, i.e. v = v i for any i A k. Since (p, X) is envy-free, we have using Claim 4.1 that u i (p, X) = 0 for all i A k, hence j J vq j p j = 0. Suppose some j J does not satisfy vq j p j = 0. Arrange all j J in descending order of vq j p j. Any proper prefix P of this sequence satisfies j P vq j p j > 0. Then buyer ī envies this prefix Maximizing Revenue for a Given Set of Winners and Allocated Items Suppose that S is a candidate winner set and T is a subset of items, where T = d(s). We want to know if there is an envy-free solution such that S is the set of winners and S wins items in T ; if yes, we want to find one that maximizes revenue. This problem can be solved easily by a linear program with an exponential number of constraints for each i S. The following combinatorial algorithm does the same thing; the idea inside is critical to our main algorithm. We will use the following notations: S = {i 1, i 2,..., i t } with i 1 < i 2 < < i t and T = {j 1, j 2,..., j l } with j 1 < j 2 < < j l. Let i b(s) be the winner of j s, s = 1, 2,..., l. Remark 4.1. It should be noted that in LP (k), the objective function is equivalent to maximize p k. Also note that d i = O(1) for constraint (5) of MaxRevenue, for any i [n]. By the pricing rule (2),(6) and (c) of MaxRevenue(S, T ), the total revenue j T p j obtained is a linear increasing function of p k, hence maximizing p k is equivalent to maximizing the total revenue. We establish the following properties: Proposition 4.5. Let (p, X) be computed in terms of LP (k ) where k X it in MaxRevenue(S, T ). Let i b(u) be the winner of j u. Use the convention j l dit +1 = k. For s = 1, 2,..., l d it, we have 1. v ib(s) q js+1 p js+1 0; 2. p js q js p j s+1 q js+1 ; 3. p ji p ji+1. 15

17 MaxRevenue(S, T ): T = d(s) Let L S = FindLoser(S). Allocation X Input a candidate winner set S and a subset of items T where Let X i, for each buyer i / S. Allocate items in T to buyers in S according to the following rule (by Lemma 2.3): Buyers with smaller indices obtain items with smaller indices. Price p Let Y = For each item j / T, let p j =. For each item k X it, do the following (a) Let J be the last 2 items with the largest indices in T if T 2 and J = T otherwise. Run the following linear program (denoted by LP (k) ), which computes prices for items in X it 1 X it min v it 1 q k p k s.t. v it 1 q k p k v it 1 q j p j j X it (1) j X it (v it q j p j ) = 0 (2) v it 1 q j p j = v it 1 q k p k j X it 1 (3) v it q j p j v it q j p j j X it 1, j X it (4) j J (v iq j p j ) 0 i L S, J J with J = d i (5) p js = v b(s) (q js q js+1 ) + p js+1 j s J X it X it 1 (6) (b) If the LP (k) in (a) has a feasible solution, let Y Y {k}. (c) For each item j s X i1 X it 2 in the reverse order let p js = v ib(s) (q js q js+1 ) + p js+1 (d) Denote the price vector computed above by p (k). If Y =, return that there is no price vector p such that (p, X) is envy-free. Otherwise, Let k Y have the largest total revenue for which (p (k ), X) is an envy-free solution. Output the tuple (p (k ), X). Proof. For the first inequality, consider the last case, v it 1 q k p k 0. Assume it does not hold. By (1) in Algorithm MaxRevenue, j X it (v it 1 q j p j ) < 0. Therefore, j X it (v it q j p j ) < 0, which contradicts Formula (2). Further, v iu q k p k 0 for all u : 1 u t 1. That is, all other buyers have nonnegative utility on item k. Now consider s = 1, 2,..., l d it. By (6) and (c) in the algorithm, using the convention j l dit +1 = k, item 1 holds as following v ib(s) q js+1 p js+1 v ib(s+1) q js+1 p js+1 = v ib(s+1) q js+2 p js+2 v it 1 q k p k 0. 16